[LHM]

Sorry, I know this is a really long question, but I'd really really really appreciate it if someone could answer it.

The percentages of NaHCO₃ and Na₂CO₃ are to be determined in a mixture of them with KCl. A 0.500 g sample of the mixture is dissolved in 50.0 mL of deionized water and titrated with 0.115 M HCl, resulting in a pH titration curve whether the volumes of acid needed to reach each equivalence point are 9.63 mL and 34.27 mL, respectively. Determine the number of grams of Na₂CO₃ and NaHCO₃ and their percentages in the original mixture.

I have that 0.00111 mol H⁺ is needed to reach the first equivalence point and 0.00394 mol H⁺ is needed to reach the second equivalence point, which matches up with the answer key. I've also figured out that there are 0.118 g Na₂CO₃, which is 23.6% Na₂CO₃, which also matches up with the answer key, so I don't have any problems there.

However, I thought that there would be 0.00394 mol - 2(0.00111 mol) = 0.00172 mol H⁺ needed to titrate just the original amount of NaHCO₃, since 2 moles of acid are needed to titrate the original number of moles of Na₂CO₃, and this isn't included in the amount of moles needed to titrate just the original amount of NaHCO₃. But the answer key says that 0.00394 mol - (0.00111 mol) = 0.00283 mol HCl is needed to titrate the NaHCO₃ instead. Why don't you multiply the 0.00111 mol by 2 if it takes 2 moles of HCl to titrate the Na₂CO₃ to reach the 2nd equivalence point?

Thanks!

[Borek]

What do they mean by NaHCO₃? Original amount, or NaHCO₃ present AFTER reaching first end point?

[LHM]

Well I assumed they meant the original amount of NaHCO₃ because it does say to also calculate "their percentages in the original mixture." In the answer key, it shows the percentage in the original mixture as 47.6%, which they calculated using 0.00394 mol - (0.00111 mol) = 0.00283 mol HCl * 84.01 g NaHCO₃/mol.